[Leetcode] Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Follow up:
Can you solve it without using extra space?

Analysis:

分析：以上图为例，假设从开始点1到cycle开始点4的距离为h，假设fast和slow在点7碰到而且从4到7的距离为a，整个cycle的长度为c；假设fast走了n圈，slow走了m圈，fast走的是slow的两倍那么我们就有

H + n*c + a = 2 ( h + m*c + a)

==>

(n-m)c = h + a

也就是说，如果两个指针slow从7开始走，p从head开始走，那么最终他们还是会碰到7这点。因为slow和p的速度是一样的，所以，他们的4，5，6这几点肯定是重合的，那么第一个重合的点就是cycle开始的点。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode* fast = head;
        ListNode* slow = head;
        while(fast != NULL && fast->next != NULL)
        {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow)
                break;
        }
        
        if (fast == NULL || fast->next == NULL)
            return NULL;
            
        slow = head;
        while(fast != slow)
        {
            fast = fast->next;
            slow = slow->next;
        }
        
        return slow;
        
    }
};